# Thread: Card Question, random 13 cards to 4 players

1. ## Card Question, random 13 cards to 4 players

Hi,

I've been struggling with this problem for a while now. A 52 card deck is dealt out randomly to 4 players who get 13 cards each. What is the probability that one of the players receives at least two aces? I was thinking something along the lines of 52!/(13!*13!*13!*13!) but I'm not sure I'm heading in the right direction. Can someone please share some insight to this? Thanks.

2. Originally Posted by Flipz4226
Hi,

I've been struggling with this problem for a while now. A 52 card deck is dealt out randomly to 4 players who get 13 cards each. What is the probability that one of the players receives at least two aces? I was thinking something along the lines of 52!/(13!*13!*13!*13!) but I'm not sure I'm heading in the right direction. Can someone please share some insight to this? Thanks.
If no one has at least 2 aces, each player must have exactly one ace. The probability of this is

$\displaystyle \frac{4!}{4^4} = 0.09375$,

so the probability that someone has at least 2 aces is

$\displaystyle 1 - 0.09375 = 0.90625$.

3. Hmm... good idea about using the complement! Could you just please explain how you came up with $\displaystyle \frac{4!}{4^4} = 0.09375$ Thank you very much.

4. If no one has at least 2 aces, each player must have exactly one ace. The probability of this is
Why must each player have exactly one ace? Can't one player have 3 and another have 1 and the other two players have none?

5. Originally Posted by Flipz4226
Hi,

I've been struggling with this problem for a while now. A 52 card deck is dealt out randomly to 4 players who get 13 cards each. What is the probability that one of the players receives at least two aces? I was thinking something along the lines of 52!/(13!*13!*13!*13!) but I'm not sure I'm heading in the right direction. Can someone please share some insight to this? Thanks.
Originally Posted by Flipz4226
Why must each player have exactly one ace? Can't one player have 3 and another have 1 and the other two players have none?
To me, if a player has three aces then he has "at least two aces". If this is not what you intended, then you need to explain your question to us.

6. Originally Posted by Flipz4226
Hmm... good idea about using the complement! Could you just please explain how you came up with $\displaystyle \frac{4!}{4^4} = 0.09375$ Thank you very much.
Lay out the four aces and write a number from 1 to 4 above each indicating the player who receives that ace. This can be done in 4^4 ways, each of which are equally likely. Each player receives exactly one ace when the numbers are a permutation of 1-2-3-4, which can be done in 4! ways.

7. Never mind I see what you mean now. Sorry awkward